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Combination or permutation in this problem?

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Combination or permutation in this problem?

by eustudent » Sat Jun 13, 2009 2:58 pm
A certain law firm consists of 4 senior and 6 junior partners. How many different groups of 3 patners can be formed in which at least one member of the gropup is a senior partner? (Two groups are considered different if at least one group member is different)

1)48
2)100
3)120
4)288
5)600

Hey guys, can you please help me with this one. I am using both combination and permutation formula, however, I do not get to the right answer. Try your luck with this one, and i will let you know of the OA answer. I think I should apply a combination formula here.
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by rohan_vus » Sat Jun 13, 2009 3:02 pm
Should be 100
4C3 + 4C1*6C2 + 4C2*6C1 = 100
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by raleigh » Sat Jun 13, 2009 3:08 pm
This is a combination problem. The best approach is to find the total possible groups (10C3 = 120), and subtract the number of groups with no senior partners ( consists of all junior partners so 6C3=20)

The answer is 120-20 = 100, B.

The longer approach would be to find out how many groups are possible with 1 senior partner, 2 senior partners, and 3 senior partners and add these.

1 senior partner: 4C1 * 6C2 = 4 * 15 = 60
2 senior partner: 4C2 * 6C1 = 6 * 6 = 36
3 senior partner: 4C3 * 6C0 = 4 * 1 = 4

Note that those sum to 100.
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by rah_pandey » Sat Jun 13, 2009 10:34 pm
I would always advise to use the second approach for Permutation and Combination problems if one is not comfortable doing them. The first approach is neat and time saving. however they are for advanced users who can think one step ahead. Second approach is methodical and once you master the step by step approach the first approach logically follows.

Rahul
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Re: Combination or permutation in this problem?

by Vemuri » Sat Jun 13, 2009 11:21 pm
You are right. This is a combinations question. The groups are different based on the members in the group rather than how they are arranged. Let's consider Seniors as 'S' & Juniors as 'J'. The different groups are:

SJJ - 4c1*6c2 = 4*15 = 60
SSJ - 4c2*6c1 = 6*6 = 36
SSS - 4c3 = 4

So, the number of groups possible are = 60+36+4 = 100

If the groups also depend on the arrangement of the members, then the number of groups can be determined using both combinations & permutation.

SJJ - 4c1*6c2*3 = 4*15*3 = 180 (3 - number of ways the group can be arranged, i.e. SJJ, JSJ, JJS)
SSJ - 4c2*6c1*3 = 4*15*3 = 108 (3 - number of ways the group can be arranged, i.e. SSJ, SJS, JSS)
SSS = 4c3 = 4

So, the number of groups possible are = 180+108+4 = 292ways.

Hope this helps.
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